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rainbow22
Group Title
Find the limit to 0 of....
[(e^2x)1]/x
Is it possible to do this w/out a calculator?
 one year ago
 one year ago
rainbow22 Group Title
Find the limit to 0 of.... [(e^2x)1]/x Is it possible to do this w/out a calculator?
 one year ago
 one year ago

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zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
do you know l`hopital rule yet?
 one year ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.0
no... :(
 one year ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.0
You can't have a zero in the denominater. The answer is 2
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
dont just give answer, help them find answer....I dont know how to solve this wthout lhopitals rule
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
@zzr0ck3r yeah, I'm trying to find a direct way to do this, without l'Hopital's
 one year ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.0
idk, I'm just practicing for a test and this question was online...
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
you nee to know how to tke a derivative to do this, you will lear that soon
 one year ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.0
I know how to take a derivative, but how will that help me solve it?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
take the derivative of the num and the the denom. then run the limit
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
you will have (2e^2x)/1 now run the limi
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Oh, I've got one: Well we know: \[ e^x1\sim x\\ x\to 0 \] So: \[ \lim_{x\to 0}\frac{e^{2x}1}{x}=\lim_{u\to 0}\frac{e^{u}1}{\frac{1}{2}u} \]Where \(u=2x\) Due to that: \[ \lim_{x\to 0}\frac{u}{\frac{1}{2}u}=\frac{1}{\frac{1}{2}}=2 \]
 one year ago
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